An earth satellite moves in an elliptical orbit with a perigee altitude of 300 km and
an apogee altitude of 3000 km. Assume that the radius of the earth is 6378 km.
The eccentricity of the orbit is __________ (rounded off to three decimal places).
Official Answer Key:
0.168
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.55)**
**1. Intuitive Physical Setup & Given Parameters**:
An Earth satellite moves in an elliptical orbit:
- Earth radius: $R_E = 6378\text{ km}$
- Perigee altitude: $h_p = 300\text{ km}$
- Apogee altitude: $h_a = 3000\text{ km}$
We need to determine the eccentricity $e$ of the orbit, rounded off to three decimal places.
**2. Step-by-Step Orbital Mechanics Derivation**:
1. **Perigee and Apogee Radii**:
The orbital radii measured from Earth's geocenter are:
$r_p = R_E + h_p = 6378 + 300 = 6678\text{ km}$
$r_a = R_E + h_a = 6378 + 3000 = 9378\text{ km}$
2. **Formula for Orbital Eccentricity**:
For any Keplerian elliptical orbit:
$r_p = a(1 - e), \quad r_a = a(1 + e)$
Dividing the difference by the sum:
$e = \frac{r_a - r_p}{r_a + r_p}$
3. **Calculate Eccentricity**:
$r_a - r_p = 9378 - 6678 = 2700\text{ km}$
$r_a + r_p = 9378 + 6678 = 16056\text{ km}$
$e = \frac{2700}{16056} = \frac{225}{1338} \approx 0.168161$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $e \approx 0.16816$.
- **Rounding Requirement**: Three decimal places $\implies \mathbf{0.168}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.168** (Acceptable Range: **0.165 to 0.170**)
- **Exam Takeaway**: Always add Earth's radius $R_E$ to convert altitudes to geocentric radii before computing $e = \frac{r_a - r_p}{r_a + r_p}$.
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